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a) \(4x-\sqrt{x^2-4x+4}=4x-\sqrt{\left(x-2\right)^2}=4x-\left(x-2\right)=3x+2\)
b) \(3x+\sqrt{9+6x+x^2}=3x+\sqrt{\left(x+3\right)^2}=3x-\left(x+3\right)=2x-3\)
c) \(\frac{x+6\sqrt{x}+9}{x-9}=\frac{\left(\sqrt{x}+3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+3}{\sqrt{x}-3}\)
d) \(\frac{\sqrt{x^2+4x+4}}{x+2}=\frac{\sqrt{\left(x+2\right)^2}}{x+2}=\frac{\left|x+2\right|}{x+2}\)( 1 )
với x < -2 thì : \(\left(1\right)\Leftrightarrow\frac{-\left(x+2\right)}{x+2}=-1\)
với x > -2 thì : \(\left(1\right)\Leftrightarrow\frac{\left(x+2\right)}{x+2}=1\)
a,=\(4x-\sqrt{\left(x-2\right)^2}\)
=\(4x-x+2\)
=3x+2
b,\(3x+\sqrt{\left(x+3\right)^2}\)
=\(3x+x+3\)
=4x+3
a) \(x+3+\sqrt{x^2-6x+9}\left(x\le3\right)\)
\(=x+3+\sqrt{\left(x-3\right)^2}\)
\(=x+3+\left|x-3\right|\)
\(=x+3-\left(x-3\right)\)
\(=x+3-x+3\)
\(=6\)
b) \(\sqrt{x^2+4x+4}-\sqrt{x^2}\left(-2\le x\le0\right)\)
\(=\sqrt{\left(x+2\right)^2}-\sqrt{x^2}\)
\(=\left|x+2\right|-\left|x\right|\)
\(=x+2-\left(-x\right)\)
\(=x+2+x\)
\(=2x+2=2\left(x+1\right)\)
c) \(\frac{\sqrt{x^2-2x+1}}{x-1}\left(x>1\right)\)
\(=\frac{\sqrt{\left(x-1\right)^2}}{x-1}\)
\(=\frac{\left|x-1\right|}{x-1}\)
\(=\frac{x-1}{x-1}=1\)
d) \(\left|x-2\right|+\frac{\sqrt{x^2-4x+4}}{x-2}\)
\(=\left|x-2\right|+\frac{\sqrt{\left(x-2\right)^2}}{x-2}\)
\(=\left|x-2\right|+\frac{\left|x-2\right|}{x-2}\)
\(=\left|x-2\right|+\frac{-\left(x-2\right)}{x-2}\)
\(=\left|x-2\right|-1\)
\(=-\left(x-2\right)-1\)
\(=-x+2-1\)
\(=-x+1=-\left(x-1\right)\)
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGceaqabeaacaaI2a % GaeyOeI0IaaGOmaiaadIhacqGHsisldaGcaaqaaiaaiMdacqGHsisl % caaI2aGaamiEaiabgUcaRiaadIhadaahaaWcbeqaaiaaikdaaaaabe % aakmaabmaabaGaamiEaiabgYda8iaaiodaaiaawIcacaGLPaaaaeaa % cqGH9aqpcaaI2aGaeyOeI0IaaGOmaiaadIhacqGHsisldaGcaaqaam % aabmaabaGaaG4maiabgkHiTiaadIhaaiaawIcacaGLPaaadaahaaWc % beqaaiaaikdaaaaabeaaaOqaaiabg2da9iaaiAdacqGHsislcaaIYa % GaamiEaiabgkHiTmaaemaabaGaaG4maiabgkHiTiaadIhaaiaawEa7 % caGLiWoaaeaacqGH9aqpcaaI2aGaeyOeI0IaaGOmaiaadIhacqGHRa % WkcaaIZaGaeyOeI0IaamiEaaqaaiabg2da9iaaiMdacqGHsislcaaI % ZaGaamiEaaqaamaalaaabaGaaG4maiabgkHiTmaakaaabaGaamiEaa % WcbeaaaOqaaiaadIhacqGHsislcaaI5aaaamaabmaabaGaamiEaiab % gwMiZkaaicdacaGGSaGaamiEaiabgcMi5kaaiMdaaiaawIcacaGLPa % aaaeaacqGH9aqpdaWcaaqaaiabgkHiTmaabmaabaWaaOaaaeaacaWG % 4baaleqaaOGaeyOeI0IaaG4maaGaayjkaiaawMcaaaqaamaabmaaba % WaaOaaaeaacaWG4baaleqaaOGaeyOeI0IaaG4maaGaayjkaiaawMca % amaabmaabaWaaOaaaeaacaWG4baaleqaaOGaey4kaSIaaG4maaGaay % jkaiaawMcaaaaaaeaacqGH9aqpdaWcaaqaaiabgkHiTiaaigdaaeaa % daGcaaqaaiaadIhaaSqabaGccqGHRaWkcaaIZaaaaaqaamaalaaaba % GaamiEaiabgkHiTiaaiwdadaGcaaqaaiaadIhaaSqabaGccqGHRaWk % caaI2aaabaWaaOaaaeaacaWG4baaleqaaOGaeyOeI0IaaG4maaaada % qadaqaaiaadIhacqGHLjYScaaIWaGaaiilaiaadIhacqGHGjsUcaaI % 5aaacaGLOaGaayzkaaaabaGaeyypa0ZaaSaaaeaacaWG4bGaeyOeI0 % IaaGOmamaakaaabaGaamiEaaWcbeaakiabgkHiTiaaiodadaGcaaqa % aiaadIhaaSqabaGccqGHRaWkcaaI2aaabaWaaOaaaeaacaWG4baale % qaaOGaeyOeI0IaaG4maaaaaeaacqGH9aqpdaWcaaqaamaakaaabaGa % amiEaaWcbeaakmaabmaabaWaaOaaaeaacaWG4baaleqaaOGaeyOeI0 % IaaGOmaaGaayjkaiaawMcaaiabgkHiTiaaiodadaqadaqaamaakaaa % baGaamiEaaWcbeaakiabgkHiTiaaikdaaiaawIcacaGLPaaaaeaada % GcaaqaaiaadIhaaSqabaGccqGHsislcaaIZaaaaaqaaiabg2da9maa % laaabaWaaeWaaeaadaGcaaqaaiaadIhaaSqabaGccqGHsislcaaIYa % aacaGLOaGaayzkaaWaaeWaaeaadaGcaaqaaiaadIhaaSqabaGccqGH % sislcaaIZaaacaGLOaGaayzkaaaabaWaaOaaaeaacaWG4baaleqaaO % GaeyOeI0IaaG4maaaaaeaacqGH9aqpdaGcaaqaaiaadIhaaSqabaGc % cqGHsislcaaIYaaaaaa!C78C! \begin{array}{l} 6 - 2x - \sqrt {9 - 6x + {x^2}} \left( {x < 3} \right)\\ = 6 - 2x - \sqrt {{{\left( {3 - x} \right)}^2}} \\ = 6 - 2x - \left| {3 - x} \right|\\ = 6 - 2x + 3 - x\\ = 9 - 3x\\ \dfrac{{3 - \sqrt x }}{{x - 9}}\left( {x \ge 0,x \ne 9} \right)\\ = \dfrac{{ - \left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\ = \dfrac{{ - 1}}{{\sqrt x + 3}}\\ \dfrac{{x - 5\sqrt x + 6}}{{\sqrt x - 3}}\left( {x \ge 0,x \ne 9} \right)\\ = \dfrac{{x - 2\sqrt x - 3\sqrt x + 6}}{{\sqrt x - 3}}\\ = \dfrac{{\sqrt x \left( {\sqrt x - 2} \right) - 3\left( {\sqrt x - 2} \right)}}{{\sqrt x - 3}}\\ = \dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}{{\sqrt x - 3}}\\ = \sqrt x - 2 \end{array}\)
\(6-2x-\sqrt{9-6x+x^2}\)
= \(6-2x-\sqrt{\left(3-x\right)^2}\)
= \(\left\{{}\begin{matrix}6-2x-3+x\\6-2x+3-x\end{matrix}\right.\)
= \(\left\{{}\begin{matrix}3-x\\9-3x\end{matrix}\right.\)
\(\frac{3-\sqrt{x}}{x-9}\)
=\(\frac{-\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(x-3\right)}\)
= \(\frac{-1}{\sqrt{x}+3}\)
điều kiện -4<=x<=4x<=4
\(a,\sqrt{\left(x+4\right)^2}+\sqrt{\left(x-4\right)^2}\)
\(A=\left|x+4\right|+\left|x-4\right|\)
KẾT HỢP ĐIỀU KIỆN
\(A=x+4+4-x\)
\(A=8\)
\(B=\sqrt{\left(3x\right)^2-6x+1}+\sqrt{\left(2x\right)^2-12x+3^2}\)
\(B=\sqrt{\left(3x-1\right)^2}+\sqrt{\left(2x-3\right)^2}\)
\(B=\left|3x-1\right|+\left|2x-3\right|\)
\(TH1:x>=\frac{3}{2}\)
\(B=3x-1+2x-3\)
\(B=5x-4\)
\(TH2:\frac{1}{3}< =x< \frac{3}{2}\)
\(B=3x-1-2x+3\)
\(B=x+2\)
\(TH3:x< \frac{1}{3}\)
\(B=-3x+1-2x+3\)
\(B=4-5x\)
câu c và câu d tương tự
câu c tách ra: \(C=\sqrt{\left(\sqrt{x}-3\right)^2}-\sqrt{\left(2\sqrt{x}+1\right)^2}\)
còn câu d tách ra :\(D=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)
\(D=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}\)
bạn tự làm nốt câu c, d nha
a/ \(=4x-\sqrt{\left(x-2\right)^2}=4x-x+2=3x+2\)
b/ \(=3x+\sqrt{\left(x+3\right)^2}=3x+x+3=4x+3\)
c/ xem lại đb
d/ \(=\frac{\sqrt{\left(x+2\right)^2}}{x+2}=\frac{x+2}{x+2}=1\)